Demonstration of Dirac equation covariance

[Vilnius]

Demonstrations of Dirac equation covariance state:

The Dirac equation is

$(i γ^{μ} ∂_{μ} - m)ψ(x) = 0. \ \ \ \ \ \ \ \ \ \$ [1]

If coordinates change in a way that

$x \rightarrow x' = Lx, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ [2]

where $L$ is a Lorentz transformation, [1] should mantain its form, obtaining in the new system:

$(i γ^{μ} ∂'_{μ} - m)ψ'(x') = 0, \ \ \ \ \ \ \ \$ [3]


where

$ψ'(x') = S(L)ψ(x) \ \ \ \ \ \ \ \$ [4]

and $S(L)$ is an invertible matrix rappresenting the fact $ψ'(x')$ should be a linear combination of $ψ(x)$ and should depend on $L$.

Remembering that from [2] stems

$∂ \rightarrow ∂' = L^{-1}∂, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ [5]

and substituting [4] and [5] in [3] we obtain


$(i γ^{μ} L^{-1 ρ}_{μ} ∂_{ρ} - m)Sψ(x) = 0. \ \ \ \ \ \ \ \ \$ [6]

Multiplying on the left for $S^{-1}$:

$(i S^{-1} γ^{μ} L^{-1 ρ}_{μ} ∂_{ρ}S - m)ψ(x) = 0$

and because $S$ depends on $L$ that don't vary along coordinates

$(i S^{-1} γ^{μ} L^{-1 ρ}_{μ}S∂_{ρ} - m)ψ(x) = 0. \ \ \ \ \ \ \ \ \$ [7]


To obtain covariance [7] must be equals to [1] so


$S^{-1} γ^{μ} L^{-1 ρ}_{μ}S = γ^{ρ}.\ \ \ \ \ \ \ \ \$ [8]


At this point all books state that [8] is equivalent to say


$S^{-1} γ^{μ} S = L^{μ}_{ρ}γ^{ρ}.\ \ \ \ \ \ \ \ \$ [9]


This requires $S$ and $L$ to commute.


I don' understand how it comes. They are both Lorentz transformations so not necessarly commute.

Thanks

 

[akhmeteli]

To obtain covariance [7] must be equals to [1] so


$S^{-1} γ^{μ} L^{-1 ρ}_{μ}S = γ^{ρ}.\ \ \ \ \ \ \ \ \$ [8]


At this point all books state that [8] is equivalent to say


$S^{-1} γ^{μ} S = L^{μ}_{ρ}γ^{ρ}.\ \ \ \ \ \ \ \ \$ [9]


This requires $S$ and $L$ to commute.

I did not check it properly, but I suspect you just need to multiply both sides of [8] by S from the left and by $S^{-1}$ from the right and take the inverse of both sides.

 

[The_Duck]

At this point all books state that [8] is equivalent to say


$S^{-1} γ^{μ} S = L^{μ}_{ρ}γ^{ρ}.\ \ \ \ \ \ \ \ \$ [9]


This requires $S$ and $L$ to commute.


I don' understand how it comes. They are both Lorentz transformations so not necessarly commute.

Note that S and L act on different indices. The S matrices act on the suppressed spinor indices, while the L matrices act on the explicit Lorentz indices. This is why they commute. It may help to write out the spinor indices explicitly, so that everything is in terms of sums over indices instead of matrix multiplications. If you write out all the indices explictly, then everything is just an number and you can rearrange the order of terms how you want. Then you can hit both sides with an L that will cancel the L^-1 on the left and give you what you want.